Giải thích các bước giải:

a.Xét `DeltaADC,DeltaBEC` có:

`hat(ADC)=hat(BEC)=90^@;`$\hat{C}$ chung

`->`$\Delta ADC \sim \Delta BEC(g.g)$

b.Xét `DeltaAHE,DeltaBHD` có:

`hat(AEH)=hat(BDH)=90^@`

`hat(AHE)=hat(BHD)` (hai góc đối đỉnh)

`->`$\Delta AHE\sim \Delta BHD(g.g)$

`->`$\dfrac{AH}{BH}=\dfrac{HE}{HD}$

`->`$AH\cdot HD=HE\cdot BH$

c.Ta có: `ADbotBC,BEbotAC.ADnnBE=H`

`->H` là trực tâm `DeltaABC`

`->CHbotAB`

`->CFbotAB`

Xét `DeltaAFH,DeltaADB` có:

`hat(AFH)=hat(ADB)=90^@`

`hat(FAH)=hat(DAB)` (góc chung)

`->`$\Delta AFH\sim \Delta ADB(g.g)$

`->(AF)/(AD)=(AH)/(AB)`

`->AF*AB=AH*AD`

d.Ta có: 

$\dfrac{HD}{AD}=\dfrac{\dfrac12 BC\cdot HD}{\dfrac12 BC\cdot AD}=\dfrac{S_{HBC}}{S_{ABC}}\\\dfrac{HE}{BE}=\dfrac{\dfrac12 AC\cdot HE}{\dfrac12 AC\cdot BE}=\dfrac{S_{HAC}}{S_{ABC}}\\\dfrac{HF}{CF}=\dfrac{\dfrac12 AB\cdot HF}{\dfrac12 AB\cdot CF}=\dfrac{S_{HAB}}{S_{ABC}}$

`->(HD)/(AD)+(HE)/(BE)+(HF)/(CF)=(S_(HBC))/(S_(ABC))+(S_(HAC))/(S_(ABC))+(S_(HAB))/(S_(ABC))=(S_(HBC)+S_(HAC)+S_(HAB))/(S_(ABC))=(S_(ABC))/(S_(ABC))=1`